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Firing a bullet into a bock suspended by a string Help!

  • #1
Hey guys apparently the answer to this is 491 m/s, but i keep getting 34m/s by using a method whereby i find out the height reached by the block then using kinetic energy to potential energy... doesn't seem to work tho! Please help!

A 50.0 g bullet is fired into a stationary 10.0 kg block suspended on a light inextensible wire of length 1.3 m. If the bullet becomes fully embedded in the block, and the bullet-block system reaches a maximum angle of 40.0◦, find the initial velocity of the bullet.
 

Answers and Replies

  • #2
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##\delta h## is given by 1.3-1.3cos(40), and ##g\delta h = \frac{1}{2} v_i^2## (conservation of energy, where ##v_i## is the initial velocity of the block+bullet). The total momentum before and after the collision stays the same as well, so you should be able to figure out why the velocity of the bullet had a magnitude of approximately 491 m/s before the collision.
 
  • #3
Bystander
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Show your work, and it will make it easier to see what's going on.
 
  • #4
##\delta h## is given by 1.3-1.3cos(40), and ##g\delta h = \frac{1}{2} v_i^2## (conservation of energy, where ##v_i## is the initial velocity of the block+bullet). The total momentum before and after the collision stays the same as well, so you should be able to figure out why the velocity of the bullet had a magnitude of approximately 491 m/s before the collision.
I am pretty sure i did this method, and ended up with a result of 34m/s, not sure that method can work for some reason
 
  • #5
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@StonedPhysicist The method worked just fine for me. Why don't you show your working? As Bystander said, it would make things much easier.
 
  • #6
using v=√((2(m+M)gh)/m) , where m=0.05 kg M=10kg g=9.81 and h = 0.3...m i get 34m/s???
 
  • #7
694
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Why are you taking a square root for m as well?
 
  • #8
Why are you taking a square root for m as well?
i am just rearranging (m+M)gh=1/2 mv2
 
  • #9
694
114
The kinetic energy should be given by ##\frac{1}{2} (M+m) v^2##. The gain in PE of the block+bullet equals to the KE the block+bullet possessed right after collision. The equation resolves to what I previously provided. So applying conservation of momentum, we get ##mv_1=Mv_2## , so ##v_1=\frac{M}{m} \sqrt{2g\delta h}## .
 
  • #10
The kinetic energy should be given by ##\frac{1}{2} (M+m) v^2##. The gain in PE of the block+bullet equals to the KE the block+bullet possessed right after collision. The equation resolves to what I previously provided. So applying conservation of momentum, we get ##mv_1=Mv_2## , so ##v_1=\frac{M}{m} \sqrt{2g\delta h}## .
ah I see now! thankyou!!
 
  • #11
haruspex
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I am pretty sure i did this method
No you didn't. As PWiz wrote, you use the energy equation to find the speed of block+bullet just after impact. There is work loss in the impact so you cannot use energy to relate it back to the speed of the bullet before impact. You have to use conservation of momentum for that.

Edit: I see you figured it out as I was typing.
 

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